Question

Krypton-91 is a radioactive substance that decays very quickly. The function Q(t)=Qoe^-kt models radioactive decay of krypton-91. Q represents the quantity remaining after t seconds and the decay constant k is approximately 0.07. How long will it take a quantity of krypton-91 to decay t o10% of its origional amount? round your answer to the nearest second.

Answer 1

Answer:

32.894 seconds ≈ 33 seconds

Step-by-step explanation:

Function : $Q(t)=Q_0 e^{-kt}$

The function models radioactive decay of krypton-91.

The decay constant k is approximately 0.07.

$Q_0$ denotes initial amount .

We are require to find How long will it take a quantity of krypton-91 to decay to 10% of its original amount

$\Rightarrow 10\%\times Q_0$

$\Rightarrow 0.10\times Q_0$

So, we neet to find t at which $Q(t)=0.10 Q_0$

$0.10 Q_0=Q_0 e^{-0.07t}$

$0.10 =e^{-0.07t}$

Taking log both sides

$\log0.10 =-0.07t \log e$

$\log0.10 = -0.07t \times 0.434294481903$

$-1 = -0.0304006137332t$

$\frac{-1}{-0.0304006137332} = t$

$32.894= t$

Thus it will take 32.894 seconds ≈ 33 seconds a quantity of krypton-91 to decay to 10% of its original amount.

Answer 2

The "k" value should have units and should be negative.
k = -.07 / seconds
Half-Life = ln (.5) / k
Half-Life = ln (.5) / -.07
Half-Life = -.693147 / -.07
Half-Life = 9.9 seconds
(actually, wkpedia says half-life of Krypton 91 is 8.57 seconds)

elapsed time = half-life * log (bgng amt / end amt) / log (2)
elapsed time = 9.9 * log (100 / 10) / 0.30102999566
elapsed time = 9.9 * 1 / 0.30102999566
elapsed time = 9.9 * 3.3219280949
elapsed time = 32.8870881395
elapsed time = 32.89 seconds (rounded)

Source:
http://www.1728.org/halflife.htm